How do you calculate specific heat without the final temp to calculate ΔT?

You find it in your textbook...

#c_("Fe") = "0.450 J/g"^@ "C"#
#c_w = "4.184 J/g"^@ "C"#

You don't need the final temperature explicitly. Suppose both masses were #"100 g"# and both objects were the same. Then they must reach halfway between #100^@ "C"# and #20^@ "C"#, which is #60^@ "C"#.

But iron has a lower specific heat capacity than water does, so it will not withstand as much heat as water would before changing temperature.

Thus, for the same masses of #"100 g"# each, #DeltaT_("Fe") > DeltaT_("H"_2"O")#, and the final temperature would be much closer to the starting temperature of water.

If we further have #"500 g"# of water instead of #"100 g"#, it will result in a temperature even closer to the starting temperature of water than with #"100 g"# of water.

Thus, #color(blue)(20^@ "C" < T_f < 50^@ "C")#, and it will be probably #20"-"#something #""^@ "C"#.

If you want to actually calculate it... invoke conservation of energy:

#q_"Fe" + q_w = 0#

#q_"Fe" = -q_w#

Recall that #q = mcDeltaT#, with #q# being heat flow in #"J"#, #m# mass in #"g"#, #c# specific heat capacity in #"J/g"^@ "C"#, and #DeltaT# the change in temperature in #""^@ "C"#. Thus:

#m_"Fe"c_"Fe"DeltaT_"Fe" = -m_wc_wDeltaT_w#

#m_"Fe"c_"Fe"(T_f - T_"Fe") = -m_wc_w(T_f - T_w)#

Solve for the final temperature.

#m_"Fe"c_"Fe"T_f - m_"Fe"c_"Fe"T_"Fe" = -m_wc_wT_f + m_wc_wT_w#

#m_wc_wT_f + m_"Fe"c_"Fe"T_f = m_wc_wT_w + m_"Fe"c_"Fe"T_"Fe"#

The final temperature is then:

#color(blue)(T_f) = (m_wc_wT_w + m_"Fe"c_"Fe"T_"Fe")/(m_wc_w + m_"Fe"c_"Fe")#

#= ("500 g" cdot "4.184 J/g"^@ "C" cdot 20^@ "C" + "100 g" cdot "0.450 J/g"^@ "C" cdot 100^@ "C")/("500 g" cdot "4.184 J/g"^@ "C" + "100 g" cdot "0.450 J/g"^@ "C")#

#= color(blue)(21.7^@ "C")#